It is estimated that the net amount of carbon dioxide fixed by photosynthesis on the landmass of Earth is \(5.5 \times 10^{16} \mathrm{~g} / \mathrm{yr}\) of \(\mathrm{CO}_{2}\). Assume that all this carbon is converted into glucose. (a) Calculate the energy stored by photosynthesis on land per year, in kJ. (b) Calculate the average rate of conversion of solar energy into plant energy in megawatts, MW \((1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s}) .\) A large nuclear power plant produces about \(10^{3} \mathrm{MW}\). The energy of how many such nuclear power plants is equivalent to the solar energy conversion?

First, we need to find out the number of moles of \(\mathrm{CO_2}\) and calculate the moles of glucose formed using the chemical equation of photosynthesis. Chemical equation of photosynthesis: \(6\mathrm{CO_2} + 6\mathrm{H_2O} \rightarrow \mathrm{C_6H_{12}O_6(IN\ Glucose)} + 6\mathrm{O_2}\) Given, net amount of \(\mathrm{CO_2}\) fixed by photosynthesis = \(5.5 \times 10^{16}\mathrm{~g/yr}\) Molar mass of $\mathrm{CO_2} = 12 + (2 \times 16) = 44\,\mathrm{g/mol}\) Number of moles of $\mathrm{CO_2} = \frac{5.5 \times 10^{16}\, \mathrm{g}}{44\, \mathrm{g/mol}} = 1.25 \times 10^{15}\,\mathrm{mol}\) From the chemical equation, 6 moles of \(\mathrm{CO_2}\) are required to produce 1 mole of glucose. So, the number of moles of glucose produced is: Number of moles of glucose = \(\frac{1.25 \times 10^{15}\, \mathrm{mol}}{6} = 2.08 \times 10^{14}\,\mathrm{mol}\)

Now we need to determine the energy stored by photosynthesis on land per year. We know the energy per mole of glucose produced in the process of photosynthesis is 2800 kJ/mol. To find the total energy stored, we'll multiply the moles of glucose with the energy per mole. Energy stored by photosynthesis = Number of moles of glucose × Energy per mole of glucose = \(2.08 \times 10^{14}\, \mathrm{mol} \times 2800\, \mathrm{kJ/mol} = 5.82 \times 10^{17}\,\mathrm{kJ/yr}\)

To find the average rate of solar energy conversion, we'll first convert the energy stored per year in kJ to joules (1 kJ = 1000 J). Then, we'll convert it into watts (1 W = 1 J/s) by dividing the value by the total number of seconds in a year. Energy stored per year = \(5.82 \times 10^{17}\, \mathrm{kJ/yr} \times 1000\, \mathrm{J/kJ} = 5.82 \times 10^{20}\, \mathrm{J/yr}\) Number of seconds in a year = \(365 \times 24 \times 3600 = 3.15 \times 10^7\, \mathrm{s/yr}\) Average rate of solar energy conversion (in MW) = \(\frac{5.82 \times 10^{20}\, \mathrm{J/yr}}{3.15 \times 10^7\, \mathrm{s/yr}\times 10^{6}\, \mathrm{J/MW}} = 1.85 \times 10^7\, \mathrm{MW}\) Now let's compare the solar energy conversion with the energy produced by a single nuclear power plant. Given, energy produced by a nuclear power plant = \(10^3\,\mathrm{MW}\) Number of nuclear power plants equivalent to solar energy conversion = \(\frac{1.85 \times 10^7\, \mathrm{MW}}{10^3\, \mathrm{MW}} = 1.85 \times 10^4\, \mathrm{nuclear\ power\ plants}\) To summarize: (a) The energy stored by photosynthesis on land per year is \(5.82 \times 10^{17}\,\mathrm{kJ/yr}\). (b) The average rate of solar energy conversion into plant energy is \(1.85 \times 10^7\,\mathrm{MW}\), which is equivalent to the energy produced by \(1.85 \times 10^4\) nuclear power plants.